# Uniqueness of splitting field

When one defines the splitting field for an arbitrary collection of polynomials, how does one show the uniqueness of such a splitting field? (I'm guessing it is still unique.) The induction argument used for the case of a single polynomial obviously doesn't work. I'm not so keen on using transfinite induction, either. I thought of direct limit approach but since the isomorphism between splitting fields of a single polynomial is not canonical, it doesn't seem to work either.

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Splitting fields always embed as the same subset of a given algebraic closure. – Qiaochu Yuan Mar 26 '11 at 3:40

The following can be found in several places, e.g., Hungerford; it is a proof via Zorn's Lemma (which in a sense is a sort of "transfinite induction", so perhaps you won't like it either). It is Zorn's Lemma that takes care of ensuring that you can "pick" compatible isomorphism on the single polynomials and then "glue them together" to get a single isomorphism for the splitting field of the entire set.

(This is a standard way to use Zorn's Lemma: you know you can do things non-canonically "one step at a time", so you consider the set of all "(possibly only) partially completed things" and order it by "compatibility". Then apply Zorn's Lemma to get a maximal element, and since you know that you can go "one step further" if necessary, that means the maximal element must be your destination already).

Suppose $K$ and $L$ are field extensions of $F$, and $S$ is a set of nonconstant poynomials in $F[x]$ such that $K$ and $L$ are both splitting fields of $S$ over $F$ (that is, every polynomial in $S$ splits in each of $K$ and $L$, and each of $K$ and $L$ are generated over $F$ by the roots of polynomials in $S$). We want to show that there is an isomorphism between $K$ and $L$.

Let $\mathscr{S}$ be the set of all triples $(E,N,\sigma)$, with $F\subseteq E\subseteq K$, $F\subseteq N\subseteq L$, and $\sigma\colon E\to N$ a field isomorphism that restricts to the identity on $F$.

Place a partial order on $\mathscr{S}$ by saying that $(E,N,\tau)\preceq (E',N',\sigma)$ if and only if $E\subseteq E'$, $N\subseteq N'$, and $\sigma\bigm|_{E}=\tau$.

We show that $\mathscr{S}$ satisfies the hypothesis of Zorn's Lemma: that is, it is a partially ordered set (trivial) such that every chain has an upper bound. Since $(F,F,\mathrm{id}_F)\in\mathscr{S}$, the empty chain has an upper bound. Now assume that $\mathscr{C}=\{(E_i,N_i,\sigma_i)\}_{i\in I}$ is a chain in $\mathscr{S}$. Let $E=\cup E_i$, $N=\cup N_i$, and define $\sigma\colon E\to N$ as follows: if $a\in E$, then there exists $i\in I$ such that $a\in E_i$; then define $\sigma(a) = \sigma_i(a)$.

This is well-defined: if $a\in E_i$ and $a\in E_j$, $i\neq j$, then either $E_i\subseteq E_j$ or $E_j\subseteq E_i$, since $\mathscr{C}$ is a chain; in the former case, $\sigma_j|_{E_i} = \sigma_i$, so $\sigma_j(a)=\sigma_i(a)$; similarly in the latter case.

Also, $\sigma$ is a field homomorphism: given $a,b\in E$, we know that $a\in E_i$ and $b\in E_j$ for some $i$ and some $j$; since $\mathscr{C}$ is a chain, either $E_i\subseteq E_j$ or $E_j\subseteq E_i$. Either way, we can find a single $k\in I$ such that $a$ and $b$ are both in $E_k$, and then \begin{align*} \sigma(a+b) &= \sigma_k(a+b) = \sigma_k(a)+\sigma_k(b) = \sigma(a)+\sigma(b)\\ \sigma(ab) &=\sigma_k(ab) = \sigma_k(a)\sigma_k(b) = \sigma(a)\sigma(b), \end{align*} since $\sigma_k$ is a field homomorphism. The map is onto, for given any $b\in N$, there exists $i\in I$ with $b\in N_i$, and then $b\in \sigma_i(E_i)\subseteq \sigma(E)$. Thus, $\sigma$ is a field isomorphism. The definition also ensures that $\sigma$ restricts to the identity on $F$.

Thus, $(E,N,\sigma)\in\mathscr{S}$, and by construction $(E_i,N_i,\sigma_i)\preceq (E,N,\sigma)$ for all $i\in I$. Thus, $(E,N,\sigma)$ is an upper bound for $\mathscr{C}$.

Thus, $\mathscr{S}$ is a partially ordered set in which every chain has a maximal element. By Zorn's Lemma, it has maximal elements. Let $(\mathcal{K},\mathcal{L},\sigma)$ be a maximal element. We show that $\mathcal{K}=K$ and $\mathcal{L}=L$.

Let $f(x)\in S$ be any polynomial in $S$. Let $M$ be the splitting field of $f(x)$ over $\mathcal{K}$ contained in $K$, and let $N$ be the splitting field of $f(x)$ over $\mathcal{L}$ contained in $L$.

From the result for a single polynomial, we know that the isomorphism $\sigma\colon\mathcal{K}\to\mathcal{L}$ can be extended to an isomorphism $\tau\colon M\to N$. In particular, $(M,N,\tau)\in\mathscr{S}$, and $(\mathcal{K},\mathcal{L},\sigma)\preceq(M,N,\tau)$. By maximality of $(\mathcal{K},\mathcal{L},\sigma)$, it follows that $(\mathcal{K},\mathcal{L},\sigma)=(M,N,\tau)$, so in particular, $f(x)$ splits in $\mathcal{K}$ and in $\mathcal{L}$. Thus, every polynomial in $S$ splits in $\mathcal{K}$, and hence $K=\mathcal{K}$; and every polynomial in $S$ splits in $\mathcal{L}$, hence $L=\mathcal{L}$.

Therefore, $\sigma\colon K=\mathcal{K}\stackrel{\cong}{\to}\mathcal{L}=L$ is an isomorphism between $K$ and $L$.

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In other words, this is the Isomorphism Extension Theorem. – PEV Mar 26 '11 at 4:25
@PEV: Which presumably the OP cannot invoke since he's trying to prove it. (-: – Arturo Magidin Mar 26 '11 at 4:26
Thanks for such a careful answer!! It was very helpful. – ashpool Mar 26 '11 at 5:32
@ashpool: You're welcome. Here's a nice exercise: Try to prove the Axiom of Choice by using Zorn's Lemma in a similar way (consider "partially completed choice functions"). It's essentially the same argument! – Arturo Magidin Mar 26 '11 at 18:54